Given an abelian group $G$, one can form the endomorphism ring $\mbox{End}(G)$ by letting $\alpha+\beta=\alpha(x)+\beta(x)$, and $\alpha\beta=\alpha(\beta(x))$, where $\alpha$ and $\beta$ are endomorphisms. Clearly, composition distributes over addition, and addition is commutative, so $\mbox{End}(G)$ is a ring. My question is: when is $\mbox{End}(G)$ commutative? Are there a nice set of criteria, or, if there is no such nice set of criteria, is there a nice class of abelian groups with commutative endomorphism rings.

This paper indicates the difficulty of any classification of torsion-free abelian groups with commutative endomorphism rings, as Corner has shown that very large torsion-free abelian groups can have commutative endomorphism rings (while the classifications up to now have basically been "only very small ones").

Thanks! I figured that it was rather unlikely that the general case would be simple, but I am surprised how easily the finitely abelian groups with commutative endomorphism rings are classified.
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Daniel MillerAug 12 '10 at 20:18

In the finitely generated case it follows by the structure theorem that the unique possibilities are $G=\mathbb{Z}$ and $G=\mathbb{Z}/({p_1}^{e_1})\oplus \mathbb{Z}/({p_2}^{e_2})\oplus\cdots\oplus \mathbb{Z}/({p_k}^{e_k})$, where $p_1,\dots,p_k$ are distinct primes.

Edit: in a word, the unique possibility is that $G$ is cyclic, as remarked by Jack Schmidt below.